Question

如何从sql server中的逗号分隔字符串中删除重复值。不使用功能

 Declare @data varchar(max) = '34.22,768.55,34.22,123.34,12,999.0,999.0'

我的预期结果应为

34.22,768.55,123.34,12,999.0

我尝试过这个查询，但它不会从变量中删除重复项。

Declare @data varchar(max) = '34.22,768.55,34.22,123.34,12,999.0,999.0'
set @data= (select '' + cast(cast('<d>'+replace(@data, ', ',',</d><d>')+'</d>'  as xml).query('distinct-values(/d)') as varchar(max)) +'')

Answer 1

试试这个

UniqNumber
---------------------------
12,123.34,34.22,768.55,999.0

<强>结果

CREATE TABLE table1
  (
    someNumber TYPE1 
    CHECK (someNumber.TYPE1 BETWEEN 11111111 AND 9999999999999)
  );

Answer 2

请尝试此操作 -

DECLARE @x AS XML=''
Declare @finalstring varchar(max) = ''
DECLARE @Param AS VARCHAR(100) = '34.22,768.55,34.22,123.34,12,999.0,999.0'
SET @x = CAST('<A>'+ REPLACE(@Param,',','</A><A>')+ '</A>' AS XML)
select @finalstring = @finalstring + value + ',' from ( 
SELECT t.value('.', 'VARCHAR(10)') Value FROM @x.nodes('/A') AS x(t))p
GROUP BY value
PRINT SUBSTRING(@finalstring,0,LEN(@finalstring))

<强>输出

12,123.34,34.22,768.55,999.0

适用于sql 2016 +

Declare @data varchar(max) = '34.22,768.55,34.22,123.34,12,999.0,999.0'
Declare @finalstring varchar(max) = ''
select @finalstring = @finalstring + value + ',' from string_split(@data,',')
GROUP BY value
PRINT SUBSTRING(@finalstring,0,LEN(@finalstring))

<强>输出

12,123.34,34.22,768.55,999.0

Answer 3

试试这个

Declare @data varchar(max) = '34.22,768.55,34.22,123.34,12,999.0,999.0'

 ;WITH CTE
 AS
 (
    SELECT
        MyStr = SUBSTRING(@data,CHARINDEX(',',@Data)+1,LEN(@data)),
        Val = SUBSTRING(@data,1,CHARINDEX(',',@data)-1)

    UNION ALL

    SELECT
        MyStr = CASE WHEN CHARINDEX(',',MyStr)>0
                        THEN SUBSTRING(MyStr,CHARINDEX(',',MyStr)+1,LEN(MyStr))
                    ELSE NULL END,
        Val = CASE WHEN CHARINDEX(',',MyStr)>0
                        THEN SUBSTRING(MyStr,1,CHARINDEX(',',MyStr)-1)
                    ELSE MyStr END
        FROM CTE
            WHERE ISNULL(REPLACE(MyStr,',',''),'')<>''
 )
 SELECT
    Val = SUBSTRING(List,1,LEN(List)-1)
    FROM
     (
     SELECT
        DISTINCT Val+','
        FROM CTE
            WHERE ISNULL(MyStr ,'')<>''
            FOR XML PATH('')
    )Q(List)

我的结果

12,123.34,34.22,768.55,999.0

Answer 4

这是另一种简单的方法。

Declare @data Nvarchar(max) = N'34.22,768.55,34.22,123.34,12,999.0,999.0'
     , @data2 Nvarchar(max)='';
SELECT @data = N'SELECT @DATA_DIST= @DATA_DIST+VAL+'','' 
     FROM (SELECT '''+replace(@data,',',''' AS VAL UNION SELECT ''')+''')A';
EXECUTE sp_executesql @data,N'@DATA_DIST varchar(MAX) OUTPUT',@DATA_DIST=@data2 OUTPUT;
SELECT LEFT(@data2,LEN(@data2)-1);

<强>结果：

12,123.34,34.22,768.55,999.0

如何从sql server中的逗号分隔字符串中删除重复项

4 个答案: